【0】README 0.1）本文部分文字描述轉(zhuǎn)自或譯自 https://en.wikipedia.org/wiki/Simpson%27s_rule和?https://en.wikipedia.org/wiki/Numerical_integration#Methods_for_one-dimensional_integrals；旨在利用java求積分；（定積分和無窮限積分） 0.2）you can also refer to this link for source code:?https://github.com/pacosonTang/postgraduate-research/tree/master/integration 0.3）o m g. CSDN編輯器掉鏈子，無法正常顯示source code, 大家湊合著看吧。oh.
【1】求定積分 1）intro：由 wikepedia 上關(guān)于 辛普森法的intro 以及 《高等數(shù)學(xué)第6版上冊(cè)同濟(jì)版》p229 關(guān)于定積分的近似計(jì)算中提到的辛普森法，本文求定積分的方法采用了辛普森近似法； 2）下面引用《高等數(shù)學(xué)第6版上冊(cè)同濟(jì)版》p229?關(guān)于辛普森法的描述

3）計(jì)算函數(shù)定積分的源代碼如下： // compute the numeric integration. public class Integration {public Integration(){}// apply simpson rule to approximately compute the integration. public double simpsonRule(double upper, double lower, int n, Function df) {double result = 0;double unit = (upper-lower)/n;double factor1 = unit / 3;double[] x = new double[n+1];for (int i = 0; i < x.length; i++) {x[i] = lower + unit*i;}for (int i = 0; i < x.length; i++) {if(i==0 || i==x.length-1) {result += df.fun(x[i]);}else if(i%2 == 0) { // if i is even num.result += 2*df.fun(x[i]);}else { // if i is odd num. result += 4*df.fun(x[i]);}} result *= factor1;return result;}// compute the standard normal distribution integration// refer to the integration table in p382 of "probability and statistics" from ZheJiang University.public double stdGaussValue(double realUpper) {Integration integration = new Integration();double upper = 1.0;double lower = 0.0;int n = 200; // splited into 200 subintervals.// double realUpper = 0.03;if(realUpper >= 5.0) {return 1.0;}double result = integration.simpsonRule(upper, lower, n, new Function() {@Overridepublic double fun(double x) {if(x==0) {return 0;}double t = realUpper-(1-x)/x;return Math.pow(Math.E, -0.5*t*t) / (x*x); }});result /= Math.pow(2*Math.PI, 0.5);result = new BigDecimal(result).setScale(6, RoundingMode.HALF_UP).doubleValue(); // save 6 decimal places.return result;} } public class IntegrationTest { //test case.public static void main(String[] args) {Integration integration = new Integration();double result = integration.stdGaussValue(4.42);System.out.println(result);}public static void main3(String[] args) {Integration integration = new Integration();double upper = 1.0;double lower = 0.0;int n = 50;double realUpper = 0.39;double result = integration.simpsonRule(upper, lower, n, new Function() {@Overridepublic double fun(double x) {if(x==0) {return 0;}double t = realUpper-(1-x)/x;return Math.pow(Math.E, -0.5*t*t) / (x*x); }});result /= Math.pow(2*Math.PI, 0.5);result = new BigDecimal(result).setScale(4, RoundingMode.HALF_UP).doubleValue(); System.out.println(result);}public static void main2(String[] args) {Integration integration = new Integration();double upper = 1.0;double lower = 0.0;int n = 10;double result = integration.simpsonRule(upper, lower, n, new Function() {@Overridepublic double fun(double x) {return Math.pow(Math.E, -x*x/2); }});result /= Math.pow(2*Math.PI, 0.5);System.out.println(result);BigDecimal decimal = new BigDecimal(result).setScale(4, RoundingMode.HALF_UP);result = Double.valueOf(decimal.toString());System.out.println(result);}public static void main1(String[] args) {Integration integration = new Integration();double upper = 1.0;double lower = 0;int n = 10;double result = integration.simpsonRule(upper, lower, n, new Function() {@Overridepublic double fun(double x) {return 4 / (1+Math.pow(x,2.0)); }});System.out.println(result);} }

Attention） A1）以上測試用例中涉及到的積分函數(shù)來自?《高等數(shù)學(xué)第6版上冊(cè)同濟(jì)版》p230的例2； A2）定積分表達(dá)式為
【2】求無窮限積分（本文以求標(biāo)準(zhǔn)正態(tài)分布的無窮下限反常積分為例） 1）求無窮限積分是基于定積分的；如何求定積分，本文在章節(jié)【1】中已經(jīng)講了；

2）所以標(biāo)準(zhǔn)正態(tài)分布的無窮下限反常積分函數(shù)可轉(zhuǎn)化為：

3）計(jì)算標(biāo)準(zhǔn)正態(tài)分布無窮下限積分的測試用例如上所示。 Attention） A1）上述求標(biāo)準(zhǔn)正態(tài)分布無窮下限積分的代碼對(duì)realUpper 有要求，小于等于5.0；因?yàn)楫?dāng)realUpper>5的話，其value=1了； A2）需要求標(biāo)準(zhǔn)正態(tài)分布的下限積分時(shí)，強(qiáng)烈建議使用?integration.stdGaussValue() 其精度要高些。

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